A ball of mass 100 g is dropped from a height h = 10 cm on a platform fixed at the top of a vertical spring (as shown in figure). The ball stays on the platform and the platform is depressed by a distance h/2. The spring constant is _____________ Nm⁻¹.
(Use g = 10 ms⁻²)

Solution

Physics sir · 4 steps

The ball falls h and then another h/2 while squashing the spring. Count the full drop of 3h/2.

1. Pick two instants. Start when the ball is released from rest. End at the lowest point, where the platform is pushed down by h/2 and the ball is again momentarily at rest. Kinetic energy is zero at both instants.
2. Energy balance. Between these instants the ball falls a total of h + h/2 = 3h/2. All of that lost gravitational potential energy is stored in the spring, which is compressed by x = h/2.

   mg(3h/2) = 1/2k(h/2)²

3. Put in numbers. With m = 0.1 kg and h = 0.1 m, the left side is 0.1 × 10 × 0.15 = 0.15 J. The right side is 1/2k(0.05)² = 1.25 × 10⁻³ k.
4. So. k = 0.15/(1.25 × 10⁻³) = 120 N m⁻¹, so the answer is 120.

The common slip: 80. Taking the fall as only h and leaving out the extra h/2 of compression gives mgh = 0.1 J and k = 80.

The idea to keep. When something falls onto a spring, measure the height it drops all the way to the lowest point. That is the free fall plus the compression, because gravity keeps doing work while the spring is being squeezed.

Answer from the NTA final answer key. Solution by PrepWiser.